The equality $G=A^{-1}A\cup A\cup A^{-1}$

Question

Let $G$ be a finite (resp. an infinite) group, $A\subseteq G$ and put $A^{-1}:=\{ a^{-1}:a\in A\}$.

Is it true that $G=A^{-1}A\cup A\cup A^{-1}$ implies $G=A^{-1}A$?

Answer 1

Here is a counter example. Let $G = \mathbb{Z}/2 \mathbb{Z}$, then $A = \{1 + 2\mathbb{Z}\}$ satisfies $A^{-1}A = \{0 + 2 \mathbb{Z}\} \neq G$ even though $G = A^{-1} A \cup A \cup A^{-1}$.